Completing the square is one of those algebra techniques that might sound complex at first—but once you understand how it works, it opens up a whole new way to solve and explore math problems.
In this guide, you’ll learn how completing the square can help you solve quadratic equations—even when factoring doesn’t work.
From step-by-step instructions and clear examples to helpful practice problems, we'll walk you through everything you need to master completing the square method.
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Completing the square is a method we use to rewrite a quadratic expression—that’s any expression that includes \( \displaystyle x^2 \)—into a perfect square trinomial, a polynomial with three terms.
For example, take this quadratic expression:
\( \displaystyle x^2 + 6x + 5 \)
By completing the square, we turn it into a perfect square trinomial which looks like this:
\( \displaystyle (x + 3)^2 - 4 \)
Completing the square is a method we usually encounter in Algebra 1 and Algebra 2, as we prepare for even more advanced math later on.
Beyond the classroom, completing the square pops up in all kinds of real-world scenarios. It’s used in physics to calculate the path of a thrown football, in engineering to design the curved tracks of a rollercoaster, and in construction to plan square patio layouts.
You'll also find it behind the smooth arcs of animated characters in video games, the parabolic antennas that beam your internet signal, and even in figuring out how high a drone will fly before it starts coming back down.
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At Mathnasium, we believe that students should understand why each step works, not just memorize procedures. When we understand the “why,” math becomes clearer—and more enjoyable.
Let’s build this process together using the example:
\( \displaystyle x^2 + 6x + 5 = 0 \)
Our goal is to isolate the variable terms (terms with the unknowns \( \displaystyle x \)) on one side.
To do this, we’ll cancel out the constant (+5) by subtracting 5 from both sides:
\( \displaystyle x^2 + 6x + 5 - 5 = -5 \)
\( \displaystyle x^2 + 6x = -5 \)
Now, we’ll trying to turn the expression on the left \( \displaystyle (x^2 + 6x) \) into a perfect square trinomial—something we can write as \( \displaystyle (x + \text{something})^2 \).
We do this by completing the square: We take half of the coefficient of x, then square it.
In this example, the coefficient of x is 6.
Half of 6 is 3.
Then we square it:
\( \displaystyle 3^2 = 9 \)
This is the number we’ll add to both sides to keep the equation balanced.
\( \displaystyle x^2 + 6x + 9 = -5 + 9 \)
\( \displaystyle x^2 + 6x + 9 = 4 \)
Which can be also written as:
\( \displaystyle x^2 + 6x + 9 = (x + 3)(x + 3) = (x + 3)^2 \)
And that’s how we go from:
\( \displaystyle x^2 + 6x + 9 = 4 \) to\( \displaystyle (x + 3)^2 = 4 \)
We now take the square root of both sides to undo the square:
\( \displaystyle \sqrt{(x + 3)^2} = \pm \sqrt{4} \)
\( \displaystyle x + 3 = \pm 2 \)
Now subtract 3 from both sides to solve for x:
\( \displaystyle x + 3 = \pm 2 \)
So we get two solutions:
\( \displaystyle x = -3 + 2 = -1 \)
\( \displaystyle x = -3 - 2 = -5 \)
We have two final answers:
\( \displaystyle x = -1 \)
\( \displaystyle x = -5 \)
Understanding how to complete the square becomes much easier when you see it in action.
Let’s walk through a couple of examples together.
Solve \( \displaystyle x^2 + 4x + 1 = 0 \)
Let’s use completing the square to solve this equation.
Step 1: Move the constant to the other side
\( \displaystyle x^2 + 4x = -1 \)
Subtract 1 from both sides:
\( \displaystyle x^2 + 4x + 1 - 1 = 0 - 1 \)
\( \displaystyle x^2 + 4x = -1 \)
Step 2: Complete the square
Take half of the coefficient of x, which is 4:
\( \displaystyle \frac{4}{2} = 2 \) which becomes \( \displaystyle 2^2 = 4 \)
Add 4 to both sides:
\( \displaystyle x^2 + 4x + 4 = -1 + 4 \)
\( \displaystyle x^2 + 4x + 4 = 3 \)
Which can be also written as:
\( \displaystyle x^2 + 4x + 4 = (x + 2)(x + 2) = (x + 2)^2 \)
And that’s how we go from:
\( \displaystyle x^2 + 4x + 4 = 3 \) to \( \displaystyle (x + 2)^2 = 3 \)
Step 3: Take the square root of both sides
\( \displaystyle \sqrt{(x + 2)^2} = \pm \sqrt{3} \)
\( \displaystyle x + 2 = \pm \sqrt{3} \)
Step 4: Solve for x
\( \displaystyle x = -2 \pm \sqrt{3} \)
And for the final answer, we get:
\( \displaystyle x = -2 - \sqrt{3} \) or \( x = -2 + \sqrt{3} \)
Why did we do all that?
We turned a messy-looking equation into something we could solve easily by rewriting it as a squared expression. That made it possible to use square roots to solve, instead of relying on guessing or factoring.
Ready to try another? Let’s spot the pattern!
Solve \( \displaystyle x^2 - 10x + 16 = 0 \)
Step 1: Move the constant to the other side
\( \displaystyle x^2 - 10x + 16 = 0 \)
Subtract 16 from both sides:
\( \displaystyle x^2 - 10x + 16 - 16 = 0 - 16 \)
\( \displaystyle x^2 - 10x = -16 \)
Step 2: Complete the square
Take half of -10:
\( \displaystyle \frac{-10}{2} = -5 \) which becomes \( \displaystyle (-5)^2 = 25 \)
Add 25 to both sides:
\( \displaystyle x^2 - 10x + 25 = -16 + 25 \)
\( \displaystyle x^2 - 10x + 25 = 9 \) which we can write as \( \displaystyle x^2 - 10x + 25 = (x - 5)(x - 5) = (x - 5)^2 \)
And that’s how we go from:
\( \displaystyle x^2 - 10x + 25 = 9 \) to \( \displaystyle (x - 5)^2 = 9 \)
Step 3: Take the square root:
\( \displaystyle \sqrt{(x - 5)^2} = \pm \sqrt{9} \)
\( \displaystyle x - 5 = \pm 3 \)
Step 4: Solve for x
\( \displaystyle x - 5 = \pm 3 \)
\( \displaystyle x = 5 + 3 \)
\( \displaystyle x = 5 +-3 \)
\( \displaystyle x = 8 \) or \( x = 2 \)
We always move the constant first.
Then we complete the square using \( \displaystyle \left( \frac{b}{2} \right)^2 \)
We factor into a binomial square, take the square root, and solve.
Once you see the structure—move, complete, factor, solve—it becomes a repeatable process you can apply to any quadratic.
Let’s see what you’ve learned. This short quiz will help you practice completing the square and build your confidence.
Don’t worry if you get something wrong—at Mathnasium, we believe every mistake is just another step toward mastering math.
Solve these expressions using completing the square method:
\( \displaystyle x^2 + 2x - 3 = 0 \)
\( \displaystyle x^2 + 10x + 21 = 0 \)
\( \displaystyle x^2 - 6x + 9 = 0 \)
\( \displaystyle x^2 + 8x + 16 = 0 \)
\( \displaystyle x^2 + 4x + 4 = 0 \)
Once you’re done, scroll down to check your answers. No peeking!
Factoring works when a quadratic can be broken into neat whole-number factors—but not every equation works that way.
Completing the square works every time, even when factoring is tricky or impossible. It also helps us rewrite equations in vertex form and solve quadratics with decimals, fractions, or roots. So, it’s a method that gives you more flexibility and power in algebra.
Yes—when solving an equation using completing the square, it’s helpful to move the constant term to the other side first. This keeps your equation clean and lets you clearly see the variable terms you're working with.
Once you’ve completed the square, you can bring everything back together.
That’s totally okay! Completing the square often leads to answers that include square roots or fractions, especially when the coefficient of x is an odd number.
At Mathnasium, we encourage students not to worry about "messy" answers—we’ll teach you how to handle them with confidence. The key is to follow each step carefully. You’ll get the hang of it with practice.
Great question! After you’ve added your special number \( \displaystyle \left( \frac{b}{2} \right)^2 \), try factoring the expression. If it turns into a perfect square like \( \displaystyle (x + a)^2 \) you did it right!
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If you’ve given our exercises a try, check your answers below:
\( \displaystyle x = 1 \text{ or } x = -3 \)
\( \displaystyle x = 1 \text{ or } x = -3 \)
\( \displaystyle x = 3 \)
\( \displaystyle x = -4 \)
\( \displaystyle x = -2 \)
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